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kiruha [24]
3 years ago
6

a sample of sucrose was burned in a calorimeter, and 41,600 joules oh heat energy were released. which statement is true about t

he water in the calorimeter?
Chemistry
1 answer:
Gala2k [10]3 years ago
5 0

It absorbed 41,600 joules.

Explanation:

If the sample of sucrose burnt in the calorimeter releases 41600Joules of heat, then the water in the calorimeter will absorb 41600J.

 The heat lost by burning the sucrose in the calorimeter is also the heat gained by the water therein.

A calorimeter is a device by prevents heat loss to the environment in any form.

It is specially designed to minimize heat loss by conduction, convection radiation.

Therefore, since no heat is lost, the heat released by burning the sucrose is used in heating the water there.

learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

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Which of the following would have the highest buffer capacity?
Mrrafil [7]

Answer:

1.25 M HCO₃⁻ / 1.25 M CO₃²⁻

Explanation:

Buffer capacity refers to the amount of a strong acid or base required per liter of the buffer to change its pH by one. This amount is directly related to the concentration of the conjugate acid-base pair in the buffer since the buffer pair neutralizes the strong acid or base.

Thus, the highest buffer capacity is found in the solution that has the highest concentration of the conjugate acid-base pair, which is 1.25 M HCO₃⁻ / 1.25 M CO₃²⁻ .

7 0
3 years ago
the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh
Harman [31]

Answer:

\boxed{\text{10 L}}

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}

Calculation:

\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}

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Answer:

B

Explanation:

makes sense

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